Counterexample to Lie's second theorem for SO(3)

Question

Lie's second theorem says that if $G$ is a simply connected Lie group, then every isomorphism $\Phi$ of its Lie algebra $\mathfrak{g}$ lifts to an isomorphism $\phi$ of $G$, i.e. such that $d\phi_e = \Phi$ where we identify $\mathfrak{g}$ with $T_e G$, the tangent space to $G$ at the identity.

Now consider $G = \mathrm{SO}(3)$, which is not simply connected. Is there an explicit counterexample to Lie's second theorem in this case? That is, can we write down a Lie algebra isomorphism $\Phi$ of $\mathfrak{so}(3)$ which is not the differential of any isomorphism $\phi$ of $\mathrm{SO}(3)$?

I feel like, if $(\eta_1, \eta_2, \eta_3)$ is the usual basis of $\mathfrak{so}(3)$, where $[\eta_i, \eta_j] = \eta_k$ cyclically, then a map like $\Phi(\eta_1)=\eta_2$, $\Phi(\eta_2)=\eta_1$, $\Phi(\eta_3)=-\eta_3$ ought to work, but I can't figure out how to prove there is no such group isomorphism $\phi$. I've been trying to find $t_1, t_2, t_3 \in \mathbb{R}$ and a relation involving $(e^{t_1 \eta_1}, e^{t_2 \eta_2}, e^{t_3 \eta_3})$ that is not satisfied by $(e^{t_1 \eta_2}, e^{t_2 \eta_1}, e^{-t_3 \eta_3})$ but I can't come up with one.

Answer 1

See $SU(2)$ as the group of quaterninons with norm $1$. Let$$\operatorname{Im}\mathbb{H}=\{ai+bj+ck\in\mathbb{H}\,|\,a,b,c\in\mathbb{R}\}$$and, for each $q\in SU(2)$, let $\varphi(q)\colon\operatorname{Im}\mathbb{H}\longrightarrow\operatorname{Im}\mathbb{H}$ be the map defined by $\varphi(q)(v)=q^{-1}vq$. Then $\operatorname{Im}\mathbb{H}\simeq\mathbb{R}^3$ and $\varphi$ is a surjective group homomorphism from $SU(2)$ onto $SO(3,\mathbb{R})$, whose kernel is $\pm1$. Then $D\varphi_1\colon\mathfrak{su}(2)\longrightarrow\mathfrak{so}(3,\mathbb{R})$ is a Lie algebra isomorphism and its inverse is a representation of $\mathfrak{so}(3,\mathbb{R})$ in $\mathbb{C}^2$. But you cannot lift it to $SO(3,\mathbb{R})$, since $SU(2)$ and $SO(3,\mathbb{R})$ are not isomorphic.

Answer 2

There is an isomorphism $r: \text{Aut}(SU(2)) \to \text{Aut}(SO(3))$, because any automorphism of $SU(2)$ necessarily preserves the center and $SO(3) = SU(2)/Z(SU(2))$. To see that $r$ is injective, observe that if $r(\rho) = \text{Id}$, then necessarily $\rho(g) = \pm g$ for each $g$. By continuity, the sign is independent of $g$, and because $\rho$ is a homomorphism, the sign is 1, as desired.

To see that $r$ is surjective, observe that taking the derivative at the identity gives an embedding $\text{Aut}(G) \hookrightarrow \text{Aut}(\mathfrak g)$. By Lie's theorem, the map $\text{Aut}(SU(2)) \to \text{Aut}(\mathfrak{su}(2))$ is an isomorphism, and in particular we have factored this isomorphism as the composite of two injective maps. Both maps must therefore be isomorphisms.

In fact, $$\text{Aut}(SU(2)) = \text{Inn}(SU(2)) \cong \text{Inn}(SO(3)) = SO(3).$$

José's answer gives an example of a map of Lie algebras which may not be lifted to a map of Lie groups, but we see here that if you're looking to find an automorphism of $\mathfrak{so}(3)$ which does not lift to an automorphism of $SO(3)$, you are out of luck. You'll have to try something more complicated.

Answer 3

As discussed in the comments to Mike's answer, if $G$ is a classical group other than $SO(8)$, then Lie's second theorem holds. For $G = SU(n)$ or $Sp(n)$, it holds because they are simply connected. For $SO(2n+1)$ there are no outer automorphisms (because the Dynkin diagram has no non-trivial symmetries). For $SO(2n)$ with $n\geq 5$, there is a unique interesting symmetry of the Dynkin diagram, but comes from a map on $SO(2n)$ given by conjugation by $\operatorname{diag}(-1,1,1....,1)$.

The group $SO(8)$ is special because its Dynkin diagram has more symmetry than every other simple Lie group: the Dynkin diagram for $D_4$ has an order $3$ rotational symmetry.

As every automorphism of the Dynkin diagram extends to an automorphism of the Lie algebra, there is a special automorphism $t:\mathfrak{so}(8)\rightarrow \mathfrak{so}(8)$ known as the triality automorphism.

I claim that there is no automorphism $T:SO(8)\rightarrow SO(8)$ which induces $t$, that is, Lie's second theorem fails for this map.

I'll write a roadmap below; proofs can be found in this paper by Varadarajan. The linked .pdf has no "Propositions", so below, I'll write, e.g. Proposition 1 (Theorem 3) to indicate that my Proposition 1 is Theorem 3 in the paper.

Proposition 1 (Theorem 3 and Lemma 4): There are precisely two conjugacy classes $\Sigma_1, \Sigma_2$ of $Spin(7)\subseteq SO(8)$. This is precisely one conjugacy class $\Sigma_0$ of $SO(7)\subseteq SO(8)$. $\square$

Now, let $\widetilde{T}:Spin(8)\rightarrow Spin(8)$ induce $t$. (The map $\widetilde{T}$ exists by Lie's second theorem.) Since $t^3$ is the identity (up to conjugacy), so is $\widetilde{T}^3$.

Proposition 2 (Theorem 5): The lifts of each of the $\Sigma_i$ to $Spin(8)$ form three distinct conjugacy classes $\widetilde{\Sigma_i}$ in $Spin(8)$. Further, $\mathbb{Z}/3\mathbb{Z} = \{Id_{Spin(8)}, \widetilde{T}, \widetilde{T}^2\}$ acts transitively on $\{\widetilde{\Sigma_i}\}$. $\square$

With these in hand, it's easy to prove the claim above, that there is no $T:SO(8)\rightarrow SO(8)$ inducing $t$. Namely, since powers of $\widetilde{T}$ act transitively on $\{\widetilde{\Sigma_i}\}$, powers of $T$ must act transitively on $\{\Sigma_i\}$. However, the elements of $\Sigma_1$ are isomorphic to $Spin(7)$, so are simply connected, while the elements of $\Sigma_0$ are isomorphic to $SO(7)$, so aren't. Thus, there is no diffeomorphism (or even homotopy equivalence!) which can move $\Sigma_0$ to $\Sigma_1$, even if we remove the restriction that maps be homomorphisms.